41.56 grams of calcium hydroxide heptahydrate contains how many moles of associated water molecules, molecules
# of molecules of associated water molecules
# of hydrogen atoms
Answer:
Molecular weight Ca(OH)2 . 7H2O = 200 g/mol
41.56 g / 200 = 0.208 moles Ca(OH)2 . 7H2O
We have 0.208 x 7 = 1.46 moles H2O
1.46 x 6.02 x 10^23 = 8.79 x 10^23 molecules H2O
Moles H =( 0.208 x 2 )+( 1.46 x 2) = 3.34
Hydrogen atoms = 3.34 x 6.02 x 10^23 = 2.01 x10^24
41.56g of hydrate(1 mole of hydrate/200.08 g hydrate per mole of hydrate in 41.56 g
.208 moles of hydrate contain 7 moles H2O each, so there ia 1.456 moles of associated H2O molecules
1.456 moles multiplied by Aavagadro's number is 8.77e23 molecules of H2O
1.456 moles of H2O and .416 moles of OH- give 3.328 moles of H atoms which is 2.00e24atoms
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